leetcode 13. Roman to Integer

题目内容

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000
For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9. 
X can be placed before L (50) and C (100) to make 40 and 90. 
C can be placed before D (500) and M (1000) to make 400 and 900.
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example:

Input: "III"
Output: 3
Example 2:

Input: "IV"
Output: 4
Example 3:

Input: "IX"
Output: 9
Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

分析过程

• 题目归类：
  技巧题目
• 题目分析：
  对于roman来说有一定的规律，
  1. 值不大于5那么出现的1一定是1，大于5后出现1一定是给原来值减一
  2. 对于V这样的直接输出5就好。
  3. 值不大于50那么出现的10一定是10，大于50之后出现的一定是-10
     4.以此类推
• 边界分析：
  • 空值分析
  • 循环边界分析
• 方法分析：
  • 数据结构分析
  • 状态机
  • 状态转移方程
  • 最优解
• 测试用例构建

代码实现

class Solution {
    public int romanToInt(String s) {
        int ref = 0;
        for(int i = s.length()-1;i>=0;i--){
            char c = s.charAt(i);
            switch(c){
                case 'I':
                    ref=(ref>=5)?ref-1:ref+1;
                    break;
                case 'V':
                    ref+=5;
                    break;
                case 'X':
                    ref=(ref>=50)?ref-10:ref+10;
                    break;
                case 'L':
                    ref +=50;
                    break;
                case 'C':
                    ref=(ref>=500)?ref-100:ref+100;
                    break;
                case 'D':
                    ref = ref+500;
                    break;
                case 'M':
                    ref = ref+1000;
                    break;
            }
        }
        return ref;
    }
}

效率提高

这个效率已经是最高的部分了。